A small college has $800$ students, $10\%$ of which are left-handed. Suppose they take an SRS of $8$ students. Let $L=$ the number of left-handed students in the sample. Which of the following would find $P(L=2)$ ? Choose 1 answer: Choose 1 answer: (Choice A) A ${800 \choose 8}(0.10)^2(0.90)^6$ (Choice B) B ${8 \choose 2}(0.10)^6(0.90)^2$ (Choice C) C ${8 \choose 2}(0.10)^2(0.90)^6$ (Choice D) D $(0.10)^6(0.90)^2$ (Choice E) E $(0.10)^2(0.90)^6$
Probability of $2$ successes We want the probability that there are $2$ successes (left-handed students) in $8$ trials (number of students sampled), so we're going to need $6$ failures (not left-handed students) as well. The probability of each success is ${0.10}$ and the probability of each failure is $0.90}$. Since we're sampling less than $10\%$ of the population, we can assume independence and multiply probabilities to find the probability of getting $2$ successes followed by $6$ failures: $\begin{aligned} P(\text{SSFFFFFF})&=\left({0.10}\right)\left({0.10}\right)\left(0.90}\right)\left(0.90}\right)\dots\left(0.90}\right) \\\\ &=\left({0.10}\right)^2\left(0.90}\right)^6 \end{aligned}$ The binomial coefficient ${n \choose k}$ SSFFFFFF isn't the only arrangement that produces $2$ successes in $8$ trials. For instance, FFFFFFSS would also produce the desired outcome. To count how many possible arrangements there are, we use the binomial coefficient ${n \choose k}$. It tells us the number of possible arrangements for $k$ successes in $n$ trials. In this problem, we want $k=2$ successes (left-handed students) in $n=8$ trials (number of students sampled), so we should use the binomial coefficient ${8 \choose 2}$. [Tell me more about the binomial coefficient.] Putting it together Each arrangement has probability $(0.10)^2(0.90)^6$ so for our final answer we multiply this probability by the number of possible arrangements: ${8 \choose 2}(0.10)^2(0.90)^6$ The answer: ${8 \choose 2}(0.10)^2(0.90)^6$